Question

In the figure shown, the coefficient of static friction between the blocks is $0.1$, while the floor is frictionless. Find the minimum value of force $F$ (in $N$ ) to cause sliding between the blocks $A$ and $B$ of masses $5kg$ and $10kg$, respectively, when the angle of string is $53^{\circ }$ with horizontal. (Use $\sin 53^{\circ }=4/5$ and $\cos 53^{\circ }=3/5$ )

Answer 1

Answer: 25

$f_1\le \mu N_1$
or $f_1\le 5$
For minimum $F$ to cause sliding,
$a_A=a_B$ and $f=f_{\max }$
or $\frac{f_{\max }}{m_A}=\frac{F\cos \theta -f_{\max }}{m_B}$
or $\frac{5}{5}=\frac{F\left(\frac{3}{5}\right)-5}{10}$
or $F=25N$